`a)f(x)-g(x)`

`=x^3-2x^2+3x+1-(x^3+x-1)`

`=x^3-2x^2+3x+1-x^3-x+1`

`=(x^3-x^3)+(3x-x)-2x^2+2`

`=-2x^2+2x+2=0`

`b)f(x)-g(x)+h(x)=0`

`<=>-2x^2+2x+2+2x^2-1=0`

`<=>2x+1=0`

`<=>2x=-1`

`<=>x=-1/2`

Vậy `x=-1/2` thì `f(x)-g(x)+h(x)=0`

a) f(x) - g(x)=-2x²+2x+2

b) f(x) - g(x) + h(x) =2x-1=0

=> 2x=1

=>x=\(\dfrac{1}{2}\)

Ta có: f(x) + g(x) = x - 2. Cho x - 2 = 0 ⇒ x = 2. Chọn D

a: \(f\left(x\right)=2x^2-x^5+4x^3-2x+1=-x^5+4x^3+2x^2-2x+1\)

\(g\left(x\right)=4x-3x^2+8-2x^5+7x^3=-2x^5+7x^3-3x^2+4x+8\)

\(h\left(x\right)=1-2x^2+4x^3-3x^5-7x^3=-3x^5-3x^3-2x^2+1\)

f(x)+g(x)+h(x)

\(=-x^5+4x^3+2x^2-2x+1-2x^5+7x^3-3x^2+4x+8-3x^5-3x^3-2x^2+1\)

\(=-6x^5+8x^3-3x^2+2x+10\)

b: f(x)-g(x)+h(x)

\(=-x^5+4x^3+2x^2-2x+1+2x^5-7x^3+3x^2-4x-8-3x^5-3x^3-2x^2+1\)

\(=-2x^5-6x^3+3x^2-6x-6\)

c: 2f(x)+3g(x)

\(=2\left(-x^5+4x^3+2x^2-2x+1\right)+3\left(-2x^5+7x^3-3x^2+4x+8\right)\)

\(=-2x^5+8x^3+4x^2-4x+2-6x^5+21x^3-9x^2+12x+24\)

\(=-8x^5+29x^3-5x^2+8x+26\)

d: g(x)-2h(x)

=\(-2x^5+7x^3-3x^2+4x+8-2\left(-3x^5-3x^3-2x^2+1\right)\)

\(=-2x^5+7x^3-3x^2+4x+8+6x^5+6x^3+4x^2-2\)

\(=4x^5+13x^3+x^2+4x+6\)

Ta có: f(x) - g(x) = x³ - 2x² + 3x + 1 - (x³ + x - 1) = -2x² + 2x

f(x) - g(x) + h(x) = -2x² + 2x + 2x² - 1 = 2x - 1

Mà: f(x) - g(x) + h(x) = 0

⇒ 2x - 1 = 0

\(\Leftrightarrow x=\dfrac{1}{2}\)

a, \(f\left(x\right)=2x^2+6x^4-3x^3+2011\)

\(=6x^4-3x^3+2x^2+2011\)

\(g\left(x\right)=2x^3-5x^2-3x^4-2012\)

\(=-3x^4+2x^3-5x^2-2012\)

b, \(f\left(x\right)+g\left(x\right)=6x^4-3x^3+2x^2+2011-3x^4+2x^3-5x^2-2012\)

\(=\left(6x^4-3x^4\right)+\left(2x^3-3x^3\right)+\left(2x^2-5x^2\right)+\left(2011-2012\right)\)

\(=3x^4-x^3-3x^2-1\)

\(f\left(x\right)-g\left(x\right)=6x^4-3x^3+2x^2+2011-\left(-3x^4+2x^3-5x^2-2012\right)\)

\(=6x^4-3x^3+2x^2+2011+3x^4-2x^3+5x^2+2012\)

\(=\left(6x^4+3x^4\right)-\left(3x^3+2x^3\right)+\left(2x^2+5x^2\right)+\left(2011+2012\right)\)

\(=9x^4-5x^3+7x^2+4023\)

Ta có: f(x):g(x)

\(=\dfrac{3x-a}{x-1}\)

\(=\dfrac{3x-3-a+3}{x-1}\)

\(=3+\dfrac{-a+3}{x-1}\)

Để f(x):g(x) có số dư là 2 thì 3-a=2

hay a=1

a: Ta có: f(x):g(x)

\(=\dfrac{3x-a}{x-1}\)

\(=\dfrac{3x-3+3-a}{x-1}\)

\(=3+\dfrac{3-a}{x-1}\)

Để f(x):g(x) có số dư là 2 thì 3-a=2

hay a=1

Thanks

a: \(F\left(x\right)=x^5-3x^2+x^3-x^2-2x+5\)

\(=x^5+x^3-4x^2-2x+5\)

\(G\left(x\right)=x^5-x^4+x^2-3x+x^2+1\)

\(=x^5-x^4+2x^2-3x+1\)

b: Ta có: \(H\left(x\right)=F\left(x\right)+G\left(x\right)\)

\(=x^5+x^3-4x^2-2x+5+x^5-x^4+2x^2-3x+1\)

\(=2x^5-x^4+x^3-2x^2-5x+6\)